# Thread: Quadratic Equation - Nature of roots

1. ## Quadratic Equation - Nature of roots

(a)Find the value of m for which the line $y=mx-3$ is a tangent to the curve $y=x+\frac{1}{x}$ and find the x-coordinate of the point at which this tangent touches the curve.

(b) Find the value of c and of d for which $(x:-5 is the solution set of $x^2+cx

For (a), how do I actually solve the quadratic equation when the equation have a $x^{-1}$?

No idea how to solve (b)... The whole question is just like alien to me....

2. This has nothing to do with quadratic equations. This is not algebra but calculus.
You have to take the derivative $y'$ of $y$ :

$y = x + \frac{1}{x} = x + x^{-1}$

Thus, following first principles, $y' = 1 - x^{-2}$

Have you done derivatives ? Can you see how to do it now ?

3. Originally Posted by Punch
(a)Find the value of m for which the line $y=mx-3$ is a tangent to the curve $y=x+\frac{1}{x}$ and find the x-coordinate of the point at which this tangent touches the curve.

(b) Find the value of c and of d for which $(x:-5 is the solution set of $x^2cx

For (a), how do I actually solve the quadratic equation when the equation have a $x^{-1}$?

No idea how to solve (b)... The whole question is just like alien to me....
hi punch ,

for (1) , you multiply the whole thing by x so you don end up with $x^{-1}$

$mx-3=x+\frac{1}{x}$

$mx^2-3x=x^2+1$

$(m-1)x^2-3x-1=0$ **

If a line is a tangent to the curve , it touches the curve at one point .

so $b^2-4ac=0$

$(-3)^2-4(m-1)(-1)=0$

calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .

for (2) , you are given the range , so

$(x+5)(x-3)<0\Rightarrow x^2+2x-15<0$

compared this with , $x^2+cx-d<0$ ?? (i think you have a typo here)

we see that c=2 , d=15 assuming that i guess it correctly .

4. Forgive my last post, I guess there must be a way to do it with algebra ... the hard way.
Forgive also what I said in this post, lol, I must be tired.
As mathaddict said, try to multiply the equations to get rid of the fractions and revert back to the good old quadratic equation. It makes it a lot easier.

5. Originally Posted by mathaddict
hi punch ,

for (1) , you multiply the whole thing by x so you don end up with $x^{-1}$

$mx-3=x+\frac{1}{x}$

$mx^2-3x=x^2+1$

$(m-1)x^2-3x-1=0$ **

If a line is a tangent to the curve , it touches the curve at one point .

so $b^2-4ac=0$

$(-3)^2-4(m-1)(-1)=0$

calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .

for (2) , you are given the range , so

$(x+5)(x-3)<0\Rightarrow x^2+2x-15<0$

compared this with , $x^2+cx-d<0$ ?? (i think you have a typo here)

we see that c=2 , d=15 assuming that i guess it correctly .
yes sorry for my typo. Well, thanks and I understood part a fully. About part b, it is about turning the x values into equation and then expanding and comparing to find c and d... it just seems so much like any other quadratic equations thanks

6. Originally Posted by Punch
(a)Find the value of m for which the line $y=mx-3$ is a tangent to the curve $y=x+\frac{1}{x}$ and find the x-coordinate of the point at which this tangent touches the curve.

(b) Find the value of c and of d for which $(x:-5 is the solution set of $x^2+cx

For (a), how do I actually solve the quadratic equation when the equation have a $x^{-1}$?
What quadratic equation are you talking about? There is no quadratic equation here. A line crosses a graph if the equation of the line and the equation of the graph have a common factor- that is, if $mx- 3= x+ \frac{1}{x}$. The line is tangent to the curve if that is a double root. To get a quadratic equation, multiply both sides of the equation by x.

For (b), start by finding solutions to the equation $x^2+ cx= d$. Those two points must separate "<" from ">". Try a single point, say x= 0, to determine which is which.

7. Sorry. the "quadratc equation i was talking about was an equation with x to the power of -1, the question have been solved.